Hello everybody!
These two problems have been boggling me for awhile now:
#1)1/(1+1/x) = a Find all values of "a" that yield no solution for "x"
#2)(6x-a)/(x-3) = 3 Find all values of "a" that yield no solution for "x"
I'm unsure of how to solve for this type of solution, but I've tried isolating the "x" variables like so:
#1) 1/(1+1/x) = a
1) 1 = a + a/x
2) 1 - a = a/x
3) x/a = 1/(1-a)
4) x = a/(1-a)
#2) (6x-a)/(x-3) = 3
1) 6x - a = 3x - 9
2) 3x - a = -9
3) 3x = a - 9
4) x = a/3 - 3
However, I don't really know how to express "x" as a no solution value to be solved for, so I may have made errors in my steps. If someone could lend me some insight, I would be very grateful!
I see what you mean with the first problem. It's easy to discern logically. I'm still confused with the second one though, and I did type it exactly like the original problem.
My friend told me the answer was "18" after plugging in values, but I'm more concerned with a proper method to find the answer than the answer itself. Any help would be appreciated!
I'd attack this somewhat differently. I'd start by look for impermissible values in the original expression.
The first problem gives an expression for a, and from that we can deduce that $x \ne 0\ and\ x \ne - 1.$
Now solve for x. $x = \dfrac{a}{1 - a} \implies a \ne 1.$ And $0 \ne x = \dfrac{a}{1 - a} \implies a \ne 0.$
So I would answer problem 1 as a is not equal to either 0 or 1.
In the second problem, it is clear that $x \ne 3.$
Now solving for x we get $\dfrac{6x - a}{x - 3} = 3 \implies 6x - a = 3x - 9 \implies 3x = a - 9 \implies x = \dfrac{a - 9}{3}.$
$3 \ne x = \dfrac{a - 9}{3} \implies 9 \ne a - 9 \implies a \ne 18.$
Solving for a we get $a = 3x + 9,$ which raises no problems. So a is not equal to 18.
Does this help
Hi,
If you know about functions, domain and range, maybe the following will help.
1. Rephrase the problem as f(x)=1/(1+1/x); the domain of f is then all reals except x=0 and x=-1. Now what value a is not in the range of f? Answer a=0 and a=1.
2. Again rephrase the problem. f(x)=(6x-a)/(x-3) for some a. For any a, the domain of f is all reals except x=3. Now the horizontal asymptote is y=6, no matter the value of a unequal to 18. (If a=18, the graph of f is the horizontal line y=6 with a hole at x=3) So the range of f is all real numbers y except y=6 except when a=18. So the only value of a so that f(x)=3 can not be solved for x is a=18. You might want to graph f for various a values.